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Search: id:A074279
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| 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Since the last occurrence of n comes one before the first occurrence of n+1 and the former is at SUM[i=0..n](i^2) = A000330(n), we have a(A000330(n)) = a(n*(n+1)*(2n+1)/6) = n and a(1+A000330(n)) = a(1+(n*(n+1)*(2n+1)/6)) = n+1. The current sequence is, loosely, the inverse function of the square pyramidal sequence. See also: A000330 Square pyramidal numbers: 0^2+1^2+2^2+...+n^2 = n(n+1)(2n+1)/6. A000330 has many alternative formulae, thus yielding many alternative formulae for the current sequence. - Jonathan Vos Post (jvospost3(AT)gmail.com), Mar 18 2006
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LINKS
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Y.-F. S. Petermann, J.-L. Remy and I. Vardi, Discrete derivatives of sequences, Adv. in Appl. Math. 27 (2001), 562-84.
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CROSSREFS
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Cf. A000217, A000330, A006331, A050446, A050447, A000537, A006003, A005900.
Adjacent sequences: A074276 A074277 A074278 this_sequence A074280 A074281 A074282
Sequence in context: A156875 A066339 A052375 this_sequence A072750 A029835 A074280
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KEYWORD
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nonn
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AUTHOR
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Jon Perry (perry(AT)globalnet.co.uk), Sep 21 2002
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